JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 1)
If ey
+ xy = e, the ordered pair $$\left( {{{dy} \over {dx}},{{{d^2}y} \over {d{x^2}}}} \right)$$ at x = 0 is equal to :
$$\left( {{1 \over e}, - {1 \over {{e^2}}}} \right)$$
$$\left( { - {1 \over e},{1 \over {{e^2}}}} \right)$$
$$\left( { - {1 \over e}, - {1 \over {{e^2}}}} \right)$$
$$\left( {{1 \over e},{1 \over {{e^2}}}} \right)$$
Explanation
y = 1 $$ \Rightarrow $$ x = 0
$${e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0$$
$$ \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}$$
$$ \Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0$$
x = 0, y = 1
$$ \Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0$$
$$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}$$
$${e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0$$
$$ \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}$$
$$ \Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0$$
x = 0, y = 1
$$ \Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0$$
$$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}$$
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